\(\int x^3 \sqrt {a x+b x^3} \, dx\) [45]

Optimal result

Integrand size = 17, antiderivative size = 163 \[ \int x^3 \sqrt {a x+b x^3} \, dx=-\frac {20 a^2 \sqrt {a x+b x^3}}{231 b^2}+\frac {4 a x^2 \sqrt {a x+b x^3}}{77 b}+\frac {2}{11} x^4 \sqrt {a x+b x^3}+\frac {10 a^{11/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{231 b^{9/4} \sqrt {a x+b x^3}} \] Output:

-20/231*a^2*(b*x^3+a*x)^(1/2)/b^2+4/77*a*x^2*(b*x^3+a*x)^(1/2)/b+2/11*x^4* 
(b*x^3+a*x)^(1/2)+10/231*a^(11/4)*x^(1/2)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/( 
a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x^(1/2)/a^(1/ 
4)),1/2*2^(1/2))/b^(9/4)/(b*x^3+a*x)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.58 \[ \int x^3 \sqrt {a x+b x^3} \, dx=\frac {2 \sqrt {x \left (a+b x^2\right )} \left (\sqrt {1+\frac {b x^2}{a}} \left (-5 a^2+2 a b x^2+7 b^2 x^4\right )+5 a^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{77 b^2 \sqrt {1+\frac {b x^2}{a}}} \] Input:

Integrate[x^3*Sqrt[a*x + b*x^3],x]
 

Output:

(2*Sqrt[x*(a + b*x^2)]*(Sqrt[1 + (b*x^2)/a]*(-5*a^2 + 2*a*b*x^2 + 7*b^2*x^ 
4) + 5*a^2*Hypergeometric2F1[-1/2, 1/4, 5/4, -((b*x^2)/a)]))/(77*b^2*Sqrt[ 
1 + (b*x^2)/a])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1927, 1930, 1930, 1917, 266, 761}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^3 \sqrt {a x+b x^3} \, dx\)

\(\Big \downarrow \) 1927

\(\displaystyle \frac {2}{11} a \int \frac {x^4}{\sqrt {b x^3+a x}}dx+\frac {2}{11} x^4 \sqrt {a x+b x^3}\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {2}{11} a \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \int \frac {x^2}{\sqrt {b x^3+a x}}dx}{7 b}\right )+\frac {2}{11} x^4 \sqrt {a x+b x^3}\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {2}{11} a \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a \int \frac {1}{\sqrt {b x^3+a x}}dx}{3 b}\right )}{7 b}\right )+\frac {2}{11} x^4 \sqrt {a x+b x^3}\)

\(\Big \downarrow \) 1917

\(\displaystyle \frac {2}{11} a \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a \sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {x} \sqrt {b x^2+a}}dx}{3 b \sqrt {a x+b x^3}}\right )}{7 b}\right )+\frac {2}{11} x^4 \sqrt {a x+b x^3}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2}{11} a \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {2 a \sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{3 b \sqrt {a x+b x^3}}\right )}{7 b}\right )+\frac {2}{11} x^4 \sqrt {a x+b x^3}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{11} a \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a x+b x^3}}\right )}{7 b}\right )+\frac {2}{11} x^4 \sqrt {a x+b x^3}\)

Input:

Int[x^3*Sqrt[a*x + b*x^3],x]
 

Output:

(2*x^4*Sqrt[a*x + b*x^3])/11 + (2*a*((2*x^2*Sqrt[a*x + b*x^3])/(7*b) - (5* 
a*((2*Sqrt[a*x + b*x^3])/(3*b) - (a^(3/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sq 
rt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x 
])/a^(1/4)], 1/2])/(3*b^(5/4)*Sqrt[a*x + b*x^3])))/(7*b)))/11
 

Defintions of rubi rules used

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1917

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + 
b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])   Int[ 
x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !Integ 
erQ[p] && NeQ[n, j] && PosQ[n - j]
 

rule 1927

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) 
, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
 

rule 1930

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p 
+ 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1)))   I 
nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, 
x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt 
Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
 

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {2 \left (-21 b^{2} x^{4}-6 a b \,x^{2}+10 a^{2}\right ) x \left (b \,x^{2}+a \right )}{231 b^{2} \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {10 a^{3} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{231 b^{3} \sqrt {b \,x^{3}+a x}}\)	\(158\)
default	\(\frac {2 x^{4} \sqrt {b \,x^{3}+a x}}{11}+\frac {4 a \,x^{2} \sqrt {b \,x^{3}+a x}}{77 b}-\frac {20 a^{2} \sqrt {b \,x^{3}+a x}}{231 b^{2}}+\frac {10 a^{3} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{231 b^{3} \sqrt {b \,x^{3}+a x}}\)	\(168\)
elliptic	\(\frac {2 x^{4} \sqrt {b \,x^{3}+a x}}{11}+\frac {4 a \,x^{2} \sqrt {b \,x^{3}+a x}}{77 b}-\frac {20 a^{2} \sqrt {b \,x^{3}+a x}}{231 b^{2}}+\frac {10 a^{3} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{231 b^{3} \sqrt {b \,x^{3}+a x}}\)	\(168\)

Input:

int(x^3*(b*x^3+a*x)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/231*(-21*b^2*x^4-6*a*b*x^2+10*a^2)/b^2*x*(b*x^2+a)/(x*(b*x^2+a))^(1/2)+ 
10/231*a^3/b^3*(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-2* 
(x-(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)/(b*x^ 
3+a*x)^(1/2)*EllipticF(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/ 
2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.36 \[ \int x^3 \sqrt {a x+b x^3} \, dx=\frac {2 \, {\left (10 \, a^{3} \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (21 \, b^{3} x^{4} + 6 \, a b^{2} x^{2} - 10 \, a^{2} b\right )} \sqrt {b x^{3} + a x}\right )}}{231 \, b^{3}} \] Input:

integrate(x^3*(b*x^3+a*x)^(1/2),x, algorithm="fricas")
 

Output:

2/231*(10*a^3*sqrt(b)*weierstrassPInverse(-4*a/b, 0, x) + (21*b^3*x^4 + 6* 
a*b^2*x^2 - 10*a^2*b)*sqrt(b*x^3 + a*x))/b^3
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int x^3 \sqrt {a x+b x^3} \, dx=\int x^{3} \sqrt {x \left (a + b x^{2}\right )}\, dx \] Input:

integrate(x**3*(b*x**3+a*x)**(1/2),x)
 

Output:

Integral(x**3*sqrt(x*(a + b*x**2)), x)
 

Maxima [F]

\[ \int x^3 \sqrt {a x+b x^3} \, dx=\int { \sqrt {b x^{3} + a x} x^{3} \,d x } \] Input:

integrate(x^3*(b*x^3+a*x)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*x^3 + a*x)*x^3, x)
 

Giac [F]

\[ \int x^3 \sqrt {a x+b x^3} \, dx=\int { \sqrt {b x^{3} + a x} x^{3} \,d x } \] Input:

integrate(x^3*(b*x^3+a*x)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*x^3 + a*x)*x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \sqrt {a x+b x^3} \, dx=\int x^3\,\sqrt {b\,x^3+a\,x} \,d x \] Input:

int(x^3*(a*x + b*x^3)^(1/2),x)
 

Output:

int(x^3*(a*x + b*x^3)^(1/2), x)
 

Reduce [F]

\[ \int x^3 \sqrt {a x+b x^3} \, dx=\frac {-\frac {20 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a^{2}}{231}+\frac {4 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a b \,x^{2}}{77}+\frac {2 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, b^{2} x^{4}}{11}+\frac {10 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b \,x^{3}+a x}d x \right ) a^{3}}{231}}{b^{2}} \] Input:

int(x^3*(b*x^3+a*x)^(1/2),x)
 

Output:

(2*( - 10*sqrt(x)*sqrt(a + b*x**2)*a**2 + 6*sqrt(x)*sqrt(a + b*x**2)*a*b*x 
**2 + 21*sqrt(x)*sqrt(a + b*x**2)*b**2*x**4 + 5*int((sqrt(x)*sqrt(a + b*x* 
*2))/(a*x + b*x**3),x)*a**3))/(231*b**2)